Question

Find integration of :- $$\int \cos x\log \left(\tan {\displaystyle \frac{x}{2}}\right)dx=$$
(a) $$\sin x\log \left|\tan x\right|-x+C$$
(b) $$-\sin x\log \left|\tan {\displaystyle \frac{x}{2}}\right|+x+C$$
(c) $$-\sin x\log \left|\tan {\displaystyle \frac{x}{2}}\right|-x+C$$
(d) $$\sin x\log \left|\tan {\displaystyle \frac{x}{2}}\right|-x+C$$

Answer 1

Hint: Here, we need to find the value of the given integral. We will use integration by parts and trigonometric identities to simplify the given equation and find the integral of the given function. Integration is a method by which we can find summation of discrete data.

Formula Used: We will use the following formula:
1.Using integration by parts, the integral of the product of two differentiable functions of $$x$$ can be written as $$\int uvdx=u\int vdx-\int \left({\displaystyle \frac{d\left(u\right)}{dx}}\times \int vdx\right)dx$$, where $$u$$ and $$v$$ are the differentiable functions of $$x$$.
2.The tangent of an angle $$\theta$$ can be written as $$\tan \theta ={\displaystyle \frac{\sin \theta }{\cos \theta }}$$.
3.The secant of an angle $$\theta$$ can be written as $$\sec \theta ={\displaystyle \frac{1}{\cos \theta }}$$.

Complete step-by-step answer:
We will integrate the simplified function using integration by parts.
Rewriting the equation, we get
$$\Rightarrow \int \cos x\log \left(\tan {\displaystyle \frac{x}{2}}\right)dx=\int \cos x\times \log \left(\tan {\displaystyle \frac{x}{2}}\right)dx$$
Let $$u$$ be $$\log \left(\tan {\displaystyle \frac{x}{2}}\right)$$ and $$v$$ be $$\cos x$$.
Therefore, by integrating $$\int \cos x\times \log \left(\tan {\displaystyle \frac{x}{2}}\right)dx$$ by parts, we get
$$\Rightarrow \int \cos x\times \log \left(\tan {\displaystyle \frac{x}{2}}\right)dx=\log \left(\tan {\displaystyle \frac{x}{2}}\right)\int \left(\cos x\right)dx-\int \left[{\displaystyle \frac{d\left\{\log \left(\tan {\displaystyle \frac{x}{2}}\right)\right\}}{dx}}\times \int \left(\cos x\right)dx\right]dx$$
The derivative of a function of the form $$\log \left\{f\left(x\right)\right\}$$ is $${\displaystyle \frac{1}{f\left(x\right)}}\times {\displaystyle \frac{d\left\{f\left(x\right)\right\}}{dx}}$$.
Thus, we get
$$\Rightarrow \int \cos x\times \log \left(\tan {\displaystyle \frac{x}{2}}\right)dx=\log \left(\tan {\displaystyle \frac{x}{2}}\right)\int \left(\cos x\right)dx-\int \left[{\displaystyle \frac{1}{\tan {\displaystyle \frac{x}{2}}}}\times {\displaystyle \frac{d\left(\tan {\displaystyle \frac{x}{2}}\right)}{dx}}\times \int \left(\cos x\right)dx\right]dx$$
Simplifying the derivative, we get
$$\Rightarrow \int \cos x\times \log \left(\tan {\displaystyle \frac{x}{2}}\right)dx=\log \left(\tan {\displaystyle \frac{x}{2}}\right)\int \left(\cos x\right)dx-\int \left[{\displaystyle \frac{1}{\tan {\displaystyle \frac{x}{2}}}}\times {\sec }^2{\displaystyle \frac{x}{2}}\times {\displaystyle \frac{d\left({\displaystyle \frac{x}{2}}\right)}{dx}}\times \int \left(\cos x\right)dx\right]dx$$
Therefore, we get
$$\Rightarrow \int \cos x\times \log \left(\tan {\displaystyle \frac{x}{2}}\right)dx=\log \left(\tan {\displaystyle \frac{x}{2}}\right)\int \left(\cos x\right)dx-\int \left[{\displaystyle \frac{1}{\tan {\displaystyle \frac{x}{2}}}}\times {\sec }^2{\displaystyle \frac{x}{2}}\times {\displaystyle \frac{1}{2}}\times \int \left(\cos x\right)dx\right]dx$$
The tangent of an angle $$\theta$$ can be written as $$\tan \theta ={\displaystyle \frac{\sin \theta }{\cos \theta }}$$.
The secant of an angle $$\theta$$ can be written as $$\sec \theta ={\displaystyle \frac{1}{\cos \theta }}$$.
Rewriting the terms in the parentheses in terms of sine and cosine, we get
$$\Rightarrow \int \cos x\times \log \left(\tan {\displaystyle \frac{x}{2}}\right)dx=\log \left(\tan {\displaystyle \frac{x}{2}}\right)\int \left(\cos x\right)dx-\int \left[{\displaystyle \frac{1}{{\displaystyle \frac{\sin {\displaystyle \frac{x}{2}}}{\cos {\displaystyle \frac{x}{2}}}}}}\times {\displaystyle \frac{1}{{\cos }^2{\displaystyle \frac{x}{2}}}}\times {\displaystyle \frac{1}{2}}\times \int \left(\cos x\right)dx\right]dx$$
Simplifying the expression, we get
$$\begin{array}{l}\Rightarrow \int \cos x\times \log \left(\tan {\displaystyle \frac{x}{2}}\right)dx=\log \left(\tan {\displaystyle \frac{x}{2}}\right)\int \left(\cos x\right)dx-\int \left[{\displaystyle \frac{\cos {\displaystyle \frac{x}{2}}}{\sin {\displaystyle \frac{x}{2}}}}\times {\displaystyle \frac{1}{{\cos }^2{\displaystyle \frac{x}{2}}}}\times {\displaystyle \frac{1}{2}}\times \int \left(\cos x\right)dx\right]dx\\ \Rightarrow \int \cos x\times \log \left(\tan {\displaystyle \frac{x}{2}}\right)dx=\log \left(\tan {\displaystyle \frac{x}{2}}\right)\int \left(\cos x\right)dx-\int \left[{\displaystyle \frac{1}{2}}\times {\displaystyle \frac{1}{\sin {\displaystyle \frac{x}{2}}}}\times {\displaystyle \frac{1}{\cos {\displaystyle \frac{x}{2}}}}\times \int \left(\cos x\right)dx\right]dx\end{array}$$
Therefore, we get
$$\Rightarrow \int \cos x\times \log \left(\tan {\displaystyle \frac{x}{2}}\right)dx=\log \left(\tan {\displaystyle \frac{x}{2}}\right)\int \left(\cos x\right)dx-\int \left[{\displaystyle \frac{1}{2\sin {\displaystyle \frac{x}{2}}\cos {\displaystyle \frac{x}{2}}}}\times \int \left(\cos x\right)dx\right]dx$$
Applying the trigonometric identity $$2\sin A\cos A=\sin 2A$$, we get
$$\Rightarrow \int \cos x\times \log \left(\tan {\displaystyle \frac{x}{2}}\right)dx=\log \left(\tan {\displaystyle \frac{x}{2}}\right)\int \left(\cos x\right)dx-\int \left[{\displaystyle \frac{1}{\sin x}}\times \int \left(\cos x\right)dx\right]dx$$
Now, the integral of $$\cos x$$ is $$\sin x$$.
Substituting $$\int \cos xdx=\sin x$$ in the equation, we get
$$\Rightarrow \int \cos x\times \log \left(\tan {\displaystyle \frac{x}{2}}\right)dx=\log \left(\tan {\displaystyle \frac{x}{2}}\right)\sin x-\int \left[{\displaystyle \frac{1}{\sin x}}\times \sin x\right]dx$$
Simplifying the expression, we get
$$\Rightarrow \int \cos x\times \log \left(\tan {\displaystyle \frac{x}{2}}\right)dx=\sin x\log \left(\tan {\displaystyle \frac{x}{2}}\right)-\int \left(1\right)dx$$
The integral of a constant with respect to a variable is the product of the constant and the variable.
Therefore, we get
$$\Rightarrow \int \cos x\log \left(\tan {\displaystyle \frac{x}{2}}\right)dx=\sin x\log \left(\tan {\displaystyle \frac{x}{2}}\right)-x+C$$, where $$C$$ is the constant of integration
Therefore, we get the value of the integral $$\int \cos x\log \left(\tan {\displaystyle \frac{x}{2}}\right)dx$$ as $$\sin x\log \left(\tan {\displaystyle \frac{x}{2}}\right)-x+C$$, where $$C$$ is the constant of integration.
Thus, the correct option is option (d).

Note: We simplified $${\displaystyle \frac{d\left(\tan {\displaystyle \frac{x}{2}}\right)}{dx}}$$ as $${\sec }^2{\displaystyle \frac{x}{2}}\times {\displaystyle \frac{d\left({\displaystyle \frac{x}{2}}\right)}{dx}}$$. This is because the derivative of a function of the form $$\tan \left\{f\left(x\right)\right\}$$ is $${\sec }^2\left\{f\left(x\right)\right\}\times {\displaystyle \frac{d\left\{f\left(x\right)\right\}}{dx}}$$.
We simplified $${\displaystyle \frac{d\left({\displaystyle \frac{x}{2}}\right)}{dx}}$$ as $${\displaystyle \frac{1}{2}}$$. This is because the derivative of a function of the form $$af\left(x\right)$$ is $$a{\displaystyle \frac{d\left\{f\left(x\right)\right\}}{dx}}$$, and the derivative of a variable with respect to itself is 1.
We used the trigonometric identity $$2\sin A\cos A=\sin 2A$$. This is the trigonometric identity for the sine of the double of an angle.